(2a)
PLS DRAW A TABLE WITH, |TEST| |OBSERVATION|
|INFERENCE| AND PUT THE FOLLOWING:
|Test |: C+10cm power3 distilled water
|Observation|: C dissclved in water to give a clear
solution
|Inference|:soluble salts present
PLS DRAW A TABLE WITH, |TEST| |OBSERVATION|
|INFERENCE| AND PUT THE FOLLOWING:
|Test |: C+10cm power3 distilled water
|Observation|: C dissclved in water to give a clear
solution
|Inference|:soluble salts present
(2ai) |Test|: solution of C+ drops of
AgNO3(aq)+HNO3(aq) +excess NH3(aq)
|Observation| white precipitate occur in drops.
The precipitate dissolved in excess NH3(aq)
|Inference| Cl power- present, Cl^-1 confirmed.
AgNO3(aq)+HNO3(aq) +excess NH3(aq)
|Observation| white precipitate occur in drops.
The precipitate dissolved in excess NH3(aq)
|Inference| Cl power- present, Cl^-1 confirmed.
(2aii) |TEST| solution of C +dil Hcl+Bacl2 |OBSERVATION| white precipitate formed |INFERENCE| (SO4)^2-, (SO3)^2- or (CO3)^2- present.
(2aiii) |TEST| C+ NaOH(aq) in drops and excess |OBSERVATION| A white gelatineous ppt is formed which is soluble in excess NaOH(aq) |INFERENCE| (Zn)^2+, (Pb)^2+ or (Al)^3+ present
(2aiv) |TEST|: Solution of C+NH3(aq) in drops and then in exess |OBSERVATION|: White precipitate occur which disssolve in excess. |INFERENCE|: (Zn)^2+ confirmed.
(3)
(i ) P is delivery tube
Q is drying tower
R is quick lime
S is dry ammonia
T is retort stand
(ii)By upward delivery
(iii)It serves as a drying agent
(iv)Ca(OH)2(s)+2NH4(s)->CaCl2+2H2O
(iv)Ca(OH)2(s)+2NH4(s)->CaCl2+2H2O
(l)+2NH3(g)
(1a)
NOTE THAT ^ MEANS Raise to power.
PLS DRAW UR TABLE AS USUAL AND INPUT
THE FOLLOWING:
Burette reading: final burette reading
(cm^3)
Initial burette reading (cm^3)
Volume of acid used(cm^3)
|Rough|:22.40,0.0 ,22.40,
|1st|:22.00,0.00,22.00
|2nd|:22.00,0.00,22.00
|3rd|:22.00,0.00,22.00
Average titre value=22.00+22.00+22.
00/3 =22.00cm^3
NOTE THAT ^ MEANS Raise to power.
PLS DRAW UR TABLE AS USUAL AND INPUT
THE FOLLOWING:
Burette reading: final burette reading
(cm^3)
Initial burette reading (cm^3)
Volume of acid used(cm^3)
|Rough|:22.40,0.0 ,22.40,
|1st|:22.00,0.00,22.00
|2nd|:22.00,0.00,22.00
|3rd|:22.00,0.00,22.00
Average titre value=22.00+22.00+22.
00/3 =22.00cm^3
(1bi) 2.45=250cm root 3
Xg= 1000cm root 3
Xg = 1000x2.45/250 =2450/250
Xg = 9.8gdm^-3
Cb=conc. Of B in moldm^-3
Molar mass of Na2CO3=23*2+12+3*16 =
106g
Cb= concentration in gdm^-3/molar
mass =9.8/106=0.09245 moldm^ -3
Concentration of B in moldm^3 = 0.0925
moldm -3.
Xg= 1000cm root 3
Xg = 1000x2.45/250 =2450/250
Xg = 9.8gdm^-3
Cb=conc. Of B in moldm^-3
Molar mass of Na2CO3=23*2+12+3*16 =
106g
Cb= concentration in gdm^-3/molar
mass =9.8/106=0.09245 moldm^ -3
Concentration of B in moldm^3 = 0.0925
moldm -3.
(1bii)CAVA/CBVB = na/nb
Where ca= concentration of acid in
moldm^-3
Cb =concentration of base in moldm^-3
VA= volume of acid
Vb= volume of base
Na=no of mole of acid
Nb=no of mole of base
Ca= CB*VB*na/va * nb
Ca = 0.0925*25*22.10*1
Ca= 4.625/22.10 = 0.209 moldm
Where ca= concentration of acid in
moldm^-3
Cb =concentration of base in moldm^-3
VA= volume of acid
Vb= volume of base
Na=no of mole of acid
Nb=no of mole of base
Ca= CB*VB*na/va * nb
Ca = 0.0925*25*22.10*1
Ca= 4.625/22.10 = 0.209 moldm
(1biii) molar of HCl=1+35.5=36.5g
Concentration of A in gdm^-3 =molar
mass * concentration in moldm^ -3
=36.5*0.209=7.63gdm^ -3
Concentration of A in gdm^-3 =molar
mass * concentration in moldm^ -3
=36.5*0.209=7.63gdm^ -3
(1biv) Na2CO3+2Hcl->2Nacl+H2O+CO2 106g of Na2CO3 =1*22.4dm^-3 9.8g = xdm^3 :. xdm^-3=9.8*1*22.4/106 =219.52/106 = 2.071dm^-3
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